Proofs that $g=h$ using information about surjective and injective

Question

For the following two questions, I need some help with proofs. I think I have the first one figured out, but I would like to check my result. But I am not sure where to go next with the second proof.

For the first proof I have the following:

Let $y \in B$. Then, by the definition of surjective, there exists an element $x$ in $A$ such that $f(x)=y$. But, since $g \circ f = h \circ f$, we know that $g(y)=g(f(y))=h(f(y))=h(y).$ So, $g(x)=h(x)$ and so $g=h$.

As for the second proof, I think I want to let some arbitrary element $z$ exist in $D$, and then use the definition of injective, which states that $x_{1}=x_{2}$ whenever $g(x_{1})=g(x_{2})$. But I am not sure where to go next. Thank you in advance!

Answer 1

I see what you were going for in your first proof, but there are a few mistakes. I think you meant, given $y \in B$, by surjectivity of $f$ there is some $x \in A$ such that $f(x) = y$, and using the equality $g \circ f = h \circ f$ it follows that $g(y) = g(f(x)) = h(f(x)) = h(y)$. By the arbitrary nature of $y \in B$, we deduce $g = h$.

As for your second statement, again, you want to show that $g(d) = h(d)$ for every $d \in D$. To that end, let $d \in D$ be arbitrary. From the equality $f \circ g = f \circ h$ you know that $f(g(d)) = f(h(d))$. Thus $g(d),h(d)$ are elements of $A$ that have the same image under the injective function $f$. Invoke said injectivity to immediately conclude $g(d)=h(d)$.